Open Access

Positive solutions for Neumann boundary value problems of nonlinear second-order integro-differential equations in ordered Banach spaces

Journal of Inequalities and Applications20112011:73

https://doi.org/10.1186/1029-242X-2011-73

Received: 8 November 2010

Accepted: 30 September 2011

Published: 30 September 2011

Abstract

The paper deals with the existence of positive solutions for Neumann boundary value problems of nonlinear second-order integro-differential equations

- u ( t ) + M u ( t ) = f ( t , u ( t ) , ( S u ) ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = θ

and

u ( t ) + M u ( t ) = f ( t , u ( t ) , ( S u ) ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = θ

in an ordered Banach space E with positive cone K, where M > 0 is a constant, f : [0, 1] × K × KK is continuous, S : C([0, 1], K) → C([0, 1], K) is a Fredholm integral operator with positive kernel. Under more general order conditions and measure of noncompactness conditions on the nonlinear term f, criteria on existence of positive solutions are obtained. The argument is based on the fixed point index theory of condensing mapping in cones.

Mathematics Subject Classification (2000): 34B15; 34G20.

Keywords

nonlinear second-order integro-differential equationNeumann boundary value problempositive solutioncondensing mappingfixed point index theorem

1 Introduction

Let E be an ordered Banach space, whose positive cone K is normal with a normal constant N0, that is, if θxy, then ||x|| ≤ N0||y||, where θ is the zero element in E. We consider the existence of positive solutions for nonlinear second-order integro-differential equations
- u ( t ) + M u ( t ) = f ( t , u ( t ) , ( S u ) ( t ) ) , 0 < t < 1
(1)
and
u ( t ) + M u ( t ) = f ( t , u ( t ) , ( S u ) ( t ) ) , 0 < t < 1
(2)
satisfying Neumann boundary conditions
u ( 0 ) = u ( 1 ) = θ ,
(3)
where M > 0 is a constant, f : I × K × KK is continuous, I = [0, 1], and
( S u ) ( t ) = 0 1 D ( t , s ) u ( s ) d s

is a Fredholm integral operator with integral kernel D C(I × I, +). For convenience, we denote by D ¯ : = max ( t , s ) I × I D ( t , s ) , and by D : = min ( t , s ) I × I D ( t , s ) . In the following discussions, we always assume that D ¯ > 0 .

The existence of positive solutions for ordinary differential equations with certain boundary conditions has been studied by many authors, see [15] and the references therein. At first, Guo and Lakshmikantham in [1] discussed the existence of positive solutions for two-point boundary value problem(BVP)
- u ( t ) = f ( t , u ( t ) ) , t I , u ( 0 ) = u ( 1 ) = 0
(4)

in a Banach space E, where f : I × KK is continuous. By using cone expansion and compression fixed point theorem of condensing mapping, they proved that, if the nonlinear term f satisfies the measure of noncompactness condition

(P0) For any R > 0, f is uniformly continuous on I × K R , and there exists a constant L 0 , 1 2 such that
α ( f ( t , B ) ) L α ( B ) ,

for any t I and B K R , where K R = K B ¯ ( θ , R )

and one of the following increasing conditions:

(P2) lim x K , | | x | | 0 max t I | | f ( t , x ) | | | | x | | = 0 , and there exist 0 < β < γ < 1, ϕ K* such that ϕ(x) > 0 for any x > θ and lim x K , | | x | | + min β t γ ϕ ( f ( t , x ) ) ϕ ( x ) = + ,

(P2) ( P 2 ) lim x K , | | x | | + max t I | | f ( t , x ) | | | | x | | = 0 , and there exist 0 < β < γ < 1, ϕ K* such that ϕ(x) > 0 for any x > θ and lim x K , | | x | | 0 min β t γ ϕ ( f ( t , x ) ) ϕ ( x ) = + ,

then, the BVP(4) has at least one positive solution. Later, the same technique is employed successfully in [2] in proving the existence of positive solutions for two-point boundary value problems of ordinary differential equations in . Recently, this technique is used in [35] to investigate the existence of positive solutions for second-order ordinary differential equations
- u ( t ) + M u ( t ) = f ( t , u ( t ) ) , 0 < t < 1
(5)
and
u ( t ) + M u ( t ) = f ( t , u ( t ) ) , 0 < t < 1
(6)

with Neumann boundary conditions (3) in . Obviously, the conditions (P1) and (P2) are an extension of sup-linear condition

(P1)* f 0 : = lim x 0 + max t I f ( t , x ) x = 0 , f : = lim x + min t I f ( t , x ) x = +

and sub-linear condition

(P2)* f 0 : = lim x 0 + min t I f ( t , x ) x = + , f : = lim x + max t I f ( t , x ) x = 0

in [24] in E. On the other hand, the limits conditions (P1)* and (P2)* are equivalent to the inequality conditions (P1)** and (P2)** in :

(P1)** For any ε > 0, there exists δ > 0 such that f(t, x) ≤ εx for any 0 ≤ x < δ; For any C > 0, there exists h C+(I) such that f(t, x) ≥ Cx - h(t),

(P2)** For any C > 0, there exists δ > 0 such that f(t, x) ≥ Cx for any 0 ≤ x < δ; For any ε > 0, there exists h C+(I) such that f(t, x) ≤ εx + h(t).

Clearly, the inequality conditions (P1)** and (P2)** are more convenient to verify and apply in applications than the limits conditions (P1)* and (P2)* do.

In this paper, we will improve and extend the results in [14]. At first, by applying a new estimate of measure of noncompactness, we will delete the condition that f is uniformly continuous on I × K R in the assumption (P0), see the conditions (H0) or (H0)*. Then, more general order conditions (see conditions (H1) and (H2)) are also presented in this paper to guarantee the existence of positive solutions for the Neumann boundary value problem (1) and (3) or (2) and (3) of nonlinear second-order integro-differential equations. These order conditions are a natural extension of the inequality conditions (P1)** and (P2)** in ordered Banach spaces. The argument of the paper is based on the fixed point index theory of condensing mapping in cones.

2 Preliminaries

First, we consider the boundary value problem (1) and (3) with M > 0.

To obtain a solution of the boundary value problem (1) and (3), we require a mapping whose kernel G(t, s) is the Green's function of the boundary value problem
- u ( t ) + M u ( t ) = θ , 0 < t < 1 , u ( 0 ) = u ( 1 ) = θ .
It is known in [35] that
G ( t , s ) = cosh ( m ( 1 - t ) ) cosh ( m s ) m sinh m , 0 s t 1 , cosh ( m ( 1 - s ) ) cosh ( m t ) m sinh m , 0 t s 1 ,
where m = M , cosh x = e x + e - x 2 , sinh x = e x - e - x 2 . Furthermore, a direct calculation shows that
0 < 1 m sinh m G ( t , s ) cosh 2 m m sinh m
(7)

and 0 1 G ( t , s ) d s = 1 M .

Let (X, || · ||) be a Banach space, C(I, X) denote the Banach space of all continuous X-valued functions on interval I with the norm ||u|| C = max{||u(t)||: t I}. Let α(·) denote the Kuratowski measure of noncompactness of the bounded set in X and C(I, X). For the details of definition and properties of the measure of noncompactness, see [6]. For any B C(I, X) and t I, let B(t):= {u(t): u B} X. If B is bounded in C(I, X), then B(t) is bounded in X, and α(B(t)) ≤ α(B). Some results of α (·) are given in the following lemma.

Lemma 1[711]Let × be a Banach space. Then, we have the following results:

(1) If B C(I, X) is a bounded and equicontinuous set, then α(B(t)) is continuous on I, and
α ( B ) = max t I α ( B ( t ) ) = α ( B ( I ) ) ,
(2) If B X is a bounded set, T : XX is a linear bounded operator, then
α ( T ( B ) ) | | T | | α ( B ) ,
(3) If B = {u n } C(I, X) is a bounded and countable set, then α(B(t)) is Lebesgue integrable on I, and
α I u n ( t ) d t 2 I α ( B ( t ) ) d t ,
(4) If B X is a bounded set, then there exists a countable subset B0 B such that
α ( B ) 2 α ( B 0 ) ,
(5) If J = [a, b], u C(J, X), φ C(J, +), then
a b φ ( s ) u ( s ) d s a b φ ( s ) d s C o ¯ U ( J ) ,

where U(J):= {u(t): t J}.

Let E be an ordered Banach space, whose positive cone K is normal with a normal constant N0. We define an operator Q : C(I, K) → C(I, K) by
( Q u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , ( S u ) ( s ) ) d s , t I .
(8)

Since G(t, s) > 0 and f : I × K × KK is continuous, Q : C(I, K) → C(I, K) is continuous. Obviously, a positive solution of the boundary value problem (1) and (3) is equivalent to a nonzero fixed point of the operator Q. Next, we will use fixed point index theorem of condensing mapping in cone to seek the nonzero fixed point of Q. For this purpose, we first prove that Q : C(I, K) → C(I, K) is a condensing mapping. A mapping Q : C(I, K) → C(I, K) is said to be a condensing mapping if for any bounded set B C(I, K), we have α(Q(B)) < α(B).

Lemma 2 Assume that f C(I × K × K, K) satisfies the following condition (H0) For any R > 0, f (I × K R × K R ) is bounded, and there exist two constants L1, L2> 0 with L 1 + L 2 D ¯ < M 4 such that
α ( f ( t , B 1 , B 2 ) ) L 1 α ( B 1 ) + L 2 α ( B 2 ) ,

for any t I and B1, B2 K R , where K R is defined as in condition (P0).

Then Q : C(I, K) → C(I, K) defined by (8) is a condensing mapping.

Proof. From (8) and assumption (H0), it follows that Q maps bounded sets of C(I, K) into bounded and equicontinuous sets. Let B C(I, K) be a bounded set, we show that α(Q(B)) < α(B). Let R := sup{||u|| C + ||Su|| C : u B}, then for any t I, we have B(t) K R , (SB)(t) K R . From Lemma 1(4), there exists a countable subset B0 = {u n } B such that α(Q(B)) ≤ 2α(Q(B0)). For any t I, from Lemma 1(3) and assumption (H0), we have
α ( Q ( B 0 ) ( t ) ) = α 0 1 G ( t , s ) f ( s , u n ( s ) , ( S u n ) ( s ) ) d s : n 2 0 1 G ( t , s ) α ( { f ( s , u n ( s ) , ( S u n ) ( s ) ) : n } ) d s = 2 0 1 G ( t , s ) α ( f ( s , B 0 ( s ) , ( S B 0 ) ( s ) ) ) d s 2 0 1 G ( t , s ) ( L 1 α ( B 0 ( s ) ) + L 2 α ( ( S B 0 ) ( s ) ) ) d s .
Since SB0 is bounded and equicontinuous, by Lemma 1(1) and Lemma 1(2), we have
α ( ( S B 0 ) ( s ) ) max s I α ( ( S B 0 ) ( s ) ) = α ( S B 0 ) | | S | | α ( B 0 ) D ¯ α ( B 0 ) .
Hence, by the properties of measure of noncompactness, we have
α ( Q ( B 0 ) ( t ) ) 2 ( L 1 + L 2 D ¯ ) 0 1 G ( t , s ) d s α ( B 0 ) 2 ( L 1 + L 2 D ¯ ) M α ( B ) .
Since Q(B0) is bounded and equicontinuous, from Lemma 1(1) and assumption (H0), we have
α ( Q ( B ) ) 2 α ( Q ( B 0 ) ) = 2 max t I α ( Q ( B 0 ) ( t ) ) 4 ( L 1 + L 2 D ¯ ) M α ( B ) < α ( B ) .

This implies that Q : C(I, K) → C(I, K) is a condensing mapping. The proof is completed. □

Remark 1 Comparing with assumption (P0), assumption (H0) does not require that f is uniformly continuous on I × K R . Hence assumption (H0) is weaker than assumption (P0).

Define a cone in C(I, K) by
P : = { u C ( I , K ) : u ( t ) σ u ( τ ) , t , τ I } ,

where σ = 1 cosh 2 m . Before starting our main results, we give the following lemma.

Lemma 3 Q(C(I, K)) P.

Proof. For any u C(I, K), from (7) and (8), for any τ I, we have
( Q u ) ( τ ) cosh 2 m m sinh m 0 1 f ( s , u ( s ) , ( S u ) ( s ) ) d s .
On the other hand, for any t I, we have
( Q u ) ( t ) 1 m sinh m 0 1 f ( s , u ( s ) , ( S u ) ( s ) ) d s σ ( Q u ) ( τ ) ,

i. e. Qu P. This implies that Q(C(I, K)) P. The proof is completed. □

In order to use fixed point index theorem of condensing mapping in cones to seek nonzero fixed point of Q, we also need the following lemmas.

Lemma 4[12]Let × be a Banach space, P X be a cone, Ω X be a bounded open set, θ Ω, A : P Ω ¯ P be a condensing mapping. If uμAu for any u Ω ∩ P and 0 < μ ≤ 1, then i(A, P ∩ Ω, P) = 1.

Lemma 5[13]Let × be a Banach space, P X be a cone, Ω X be a bounded open set, A : P Ω ¯ P be a condensing mapping. If there exists a ν0 P\{θ } such that u - Auτν0for any u Ω ∩ P and τ ≥ 0, then i(A, P ∩ Ω, P ) = 0.

3 Main results

For convenience, for any r > 0, let P r := {u P : ||u|| C < r}. Then, ∂P r = {u P : ||u|| C = r}. Now, we are in the position to state and prove our main results.

Theorem 1 Let E be an ordered Banach space, whose positive cone K is normal. If M > 0 and f C(I × K × K, K) satisfies the assumption (H0) and one of the following conditions

(H1) (1) There exist two constants a, b > 0 with a + b D ¯ < M and δ > 0 such that
f ( t , u , v ) a u + b v ,

for any t I and u, v K δ , where K δ = K B ¯ ( θ , δ ) ,

(2) There exist two constants c, d > 0 with c + d D > M and h0 C(I, K) such that
f ( t , u , v ) c u + d v - h 0 ( t ) ,

for any t I and u, v K,

(H2) (1) There exist two constants c, d > 0 with c + d D > M and δ > 0 such that
f ( t , u , v ) c u + d v ,

for any t I and u, v K δ ,

(2) There exist two constants a, b > 0 with a + b D ¯ < M and h0 C(I, K) such that
f ( t , u , v ) a u + b v + h 0 ( t ) ,

for any t I and u, v K,

then the boundary value problem (1) and (3) has at least one positive solution.

Proof. Since f C(I × K × K, K) satisfies assumption (H0), from Lemmas 2 and 3, we know that Q : PP is a condensing mapping. Next, we will show that the opertor Q defined by (8) has at least one nonzero fixed point when f satisfies assumption (H1) or (H2).

If (H1) holds, let 0 < r < min { δ , δ D ¯ } , then for any t I and u ∂P r , we have ||u(t)|| ≤ ||u|| C = r < δ, | | ( S u ) ( t ) | | | | S u | | C D ¯ | | u | | C = D ¯ r < δ . Hence from assumption (H1)(1), we have
f ( t , u ( t ) , ( S u ) ( t ) ) a u ( t ) + b ( S u ) ( t ) .
(9)
We now prove that uμQu for any u ∂P r and 0 < μ ≤ 1. In fact, if there exist u0 ∂P r and 0 < μ0 ≤ 1 such that u0 = μ0Qu0, then by the definition of operator Q, u0(t) satisfies the equation
- u 0 ( t ) + M u 0 ( t ) = μ 0 f ( t , u 0 ( t ) , ( S u 0 ) ( t ) ) , 0 < t < 1
(10)
and Neumann boundary condition (3). Integrating on both sides of Equation (10) from 0 to 1, by (9), we have
M 0 1 u 0 ( t ) d t 0 1 f ( t , u 0 ( t ) , ( S u 0 ) ( t ) ) d t 0 1 ( a u 0 ( t ) + b ( S u 0 ) ( t ) ) d t ( a + b D ¯ ) 0 1 u 0 ( t ) d t .
Combining this inequality with, a + b D ¯ < M , it follows that 0 1 u 0 ( t ) d t θ . But from u0 C(I, K), we have that u0(t) ≥ θ for any t I, and from u0 ∂P r , we have that ||u0|| C = r. Thus, u0(t) ≥ θ and u0(t) θ. Therefore, 0 1 u 0 ( t ) d t > θ . This is a contradiction. Hence Q satisfies the hypotheses of Lemma 4 in P r . From Lemma 4, we have
i ( Q , P r , P ) = 1 .
(11)
On the other hand, let ν0e, where e K and ||e|| = 1, then ν0 is a solution of the boundary value problem (1) and (3) when f(t, u, Su) = Me. This implies that ν0 P . Next, we show that if R > 0 large enough, then u - Quτν0 for any u ∂P R and τ ≥ 0. In fact, if there exist u0 ∂P R and τ0 ≥ 0 such that u0 - Qu0 = τ0ν0, then by the definition of operator Q, u0(t) satisfies the equation
- u 0 ( t ) + M u 0 ( t ) = f ( t , u 0 ( t ) , ( S u 0 ) ( t ) ) + M τ 0 ν 0 , 0 < t < 1
(12)
and Neumann boundary condition (3). Integrating on both sides of Equation (12) from 0 to 1, by assumption (H1)(2), we have
M 0 1 u 0 ( t ) d t = 0 1 f ( t , u 0 ( t ) , ( S u 0 ) ( t ) ) d t + M τ 0 ν 0 0 1 ( c u 0 ( t ) + d ( S u 0 ) ( t ) - h 0 ( t ) ) d t ( c + d D ) 0 1 u 0 ( t ) d t - 0 1 h 0 ( t ) d t .
Consequently, we obtain that
0 1 u 0 ( t ) d t 1 c + d D - M 0 1 h 0 ( t ) d t .
(13)
On the other hand, from u0 P , we have
0 1 u 0 ( t ) d t 0 1 σ u 0 ( τ ) d t = σ u 0 ( τ ) θ .
Combining this inequality with (13), and by the normality of cone K in E, we obtain that
| | u 0 | | C N 0 | | h 0 | | C σ ( c + d D - M ) R ¯ .
(14)
Let R > max { δ , δ D ¯ , R ¯ } . Then, for any u ∂P R and τ ≥ 0, u - Quτν0. Hence hypotheses of Lemma 5 hold. By Lemma 5, we have
i ( Q , P R , P ) = 0 .
(15)
Now, by the additivity of fixed point index, (11) and (15), we have
i ( Q , P R \ P ¯ r , P ) = i ( Q , P R , P ) - i ( Q , P r , P ) = - 1 .

Therefore, Q has a fixed point u* in P R \ P ¯ r , which satisfies u*(t) ≥ σu*(τ) ≥ θ for any t, τ I. By the normality of cone K in E, we see that | | u * ( t ) | | σ N 0 | | u * ( τ ) | | > σ r N 0 > 0 , which implies that u* is a positive solution of the boundary value problem (1) and (3).

Next, we suppose that (H2) holds. Let 0 < r < min { δ , δ D ¯ } , then for any t I and u ∂P r , we have
f ( t , u ( t ) , ( S u ) ( t ) ) c u ( t ) + d ( S u ) ( t ) .
(16)
Let ν0e, where e K and ||e|| = 1. We now prove that u - Quτν0 for any u ∂P r and τ ≥ 0. In fact, if there exist u0 ∂P r and τ0 ≥ 0 such that u0 - Qu0 = τ0ν0, then u0(t) satisfies Equation (12) and Neumann boundary condition (3). From (12) and (16), it follows that
( c + d D - M ) 0 1 u 0 ( t ) d t - M τ 0 ν 0 θ .
Since 0 1 u 0 ( t ) d t > θ , we see that c + d D M , which is a contradiction. Hence by Lemma 5, we have
i ( Q , P r , P ) = 0 .
(17)
On the other hand, we show that if R > 0 large enough, then uμQu for any u ∂P R and 0 < μ ≤ 1. In fact, if there exist u0 ∂P R and 0 < μ0 ≤ 1 such that u0 = μ0Qu0, then u0(t) satisfies Equation (10) and Neumann boundary condition (3). From (10) and assumption (H2)(2), it follows that
M 0 1 u 0 ( t ) d t ( a + b D ¯ ) 0 1 u 0 ( t ) d t + 0 1 h 0 ( t ) d t .
By the proof of (14), we see that | | u 0 | | C N 0 | | h 0 | | C σ ( M - a - b D ¯ ) R ¯ . Let R > max { δ , δ D ¯ , R ¯ } , then uμQu for any u ∂P R and 0 < μ ≤ 1. therefore, by Lemma 4, we have
i ( Q , P R , P ) = 1 .
(18)
From (17) and (18), it follows that
i ( Q , P R \ P ¯ r , P ) = i ( Q , P R , P ) - i ( Q , P r , P ) = 1 .

Therefore, Q has a fixed point u* in P R \ P ¯ r , which is the positive solution of the boundary value problem (1) and (3). The proof is completed. □

Theorem 2 Let E be an ordered Banach space, whose positive cone K is normal. If M > 0 and f C(I × K × K, K) satisfies the assumption

(H0)* For any R > 0, f(I × K R × K R ) is bounded, and there exist two constants L1, L2> 0 with 2 L 1 + L 2 D ¯ < M such that
α ( f ( I × B 1 × B 2 ) ) L 1 α ( B 1 ) + L 2 α ( B 2 ) ,

for any B1, B2 K R , where K R is defined as in (P0) and the condition (H1) or (H2), then the boundary value problem (1) and (3) has at least one positive solution.

Proof. We only need to prove that Q : C(I, K) → C(I, K) is a condensing mapping. For any bounded set B C(I, K), let R := sup{||u|| C + ||Su|| C : u B}, then B(t) K R , (SB)(t) K R for any t I. For any u B and t I, from Lemma 1(5), we have
( Q u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , ( S u ) ( s ) ) d s 0 1 G ( t , s ) d s C o ¯ ( { f ( s , u ( s ) , ( S u ) ( s ) ) : s I } ) 0 1 G ( t , s ) d s C o ¯ ( f ( I × B ( I ) × ( S B ) ( I ) ) ) .
Consequently, we obtain that
Q ( B ) ( t ) 0 1 G ( t , s ) d s C o ¯ ( f ( I × B ( I ) × ( S B ) ( I ) ) ) .
From the properties of measure of noncompactness, Lemma 1(2) and assumption (H0)*, we have
α ( Q ( B ) ( t ) ) 0 1 G ( t , s ) d s α ( f ( I × B ( I ) × ( S B ) ( I ) ) ) 1 M ( L 1 α ( B ( I ) ) + L 2 α ( ( S B ) ( I ) ) ) 2 L 1 + L 2 D ¯ M α ( B ) .
From Lemma 1(1) and assumption (H0)*, we have
α ( Q ( B ) ) = max t I α ( Q ( B ) ( t ) ) 2 L 1 + L 2 D ¯ M α ( B ) < α ( B ) .

This implies that Q : C(I, K) → C(I, K) is a condensing mapping. The proof is completed. □

Remark 2 In assumption (H0)*, we replace L 1 + L 2 D ¯ < M 4 by 2 L 1 + L 2 D ¯ < M , but the condition α(f(I × B1 × B2)) ≤ L1α (B1) + L2α (B2) is stronger than condition α(f(t, B1, B2)) ≤ L1α (B1) + L2α (B2)(t I) in (H0), where B1, B2 K R . Hence the assumption (H0)* is different from assumption (H0). It is another improvement of assumption (P0).

The same technique can be carried over for the boundary value problem defined by (2) and (3) with M ( 0 , π 2 4 ) . We only need to change the Green's function to
G ( t , s ) = cos ( m ( 1 - t ) ) cos ( m s ) m sin m , 0 s t 1 , cos ( m ( 1 - s ) ) cos ( m t ) m sin m , 0 t s 1 ,
and the cone in C(I, K) to
P : = { u C ( I , K ) : u ( t ) σ u ( τ ) , t , τ I } ,

where now σ = cos2m. Similar to the proof of Theorems 1 and 2, we can obtain the following result.

Theorem 3 Let E be an ordered Banach space, whose positive cone K is normal. If < M < π 2 4 and f 2 C(I × K × K, K) satisfy the assumption (H0) or (H0)* and the condition (H1) or (H2), then the boundary value problem defined by (2) and (3) has at least one positive solution.

Remark 3 The conditions (H1) and (H2) are a natural extension of the inequality conditions (P1)** and (P2)** in ordered Banach space E. Hence, if f(t, u, v) = f(t, u), then Theorems 1, 2, and 3 improve and extend the main results in[24].

Declarations

Acknowledgements

Research supported by NNSF of China (10871160), the NSF of Gansu Province (0710RJZA103), and Project of NWNU-KJCXGC-3-47.

Authors’ Affiliations

(1)
Department of Mathematics, Northwest Normal University
(2)
Science College of Gansu Agricultural University

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© Yang and Liang; licensee Springer. 2011

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